A spring with a spring-constant 2.9 N/cm is compressed 28 cm and released. The 4

Question

A spring with a spring-constant 2.9 N/cm is compressed 28 cm and released. The 4 kg mass skids down the frictional incline of height 33cm and inclined at a 24 degree angle. The acceleration of gravity is 9.8 m/s^2. The path is friction less except for a distance of 0.5 m along the incline which has a coefficient of friction of 0.4. What is the final velocity vf of the mass? Answer in m/s.

Explanation / Answer

initial energu of the system is: kx^2/2+mgh=0.11368+12.936=13.05 J the system looses some energy during the friction path: E=F*d=u*mg*cos(24)*0.5=7.16 J --> at the bottom mass has only kinetic energy mv^2/2, which is initial energy minus energy wasted on friction = 13.05-7.16=5.89 J = mv^2/2 --> v=1.716 m/sec

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